How to shorten a line but maintain its angle? [closed]
$begingroup$
I have a line that looks like this:
to draw this line i have the following:
point A: (x,y)
point B: (x,y)
so for example, i will be given this path:
(90,20),(400,300)
and then i can draw a line because i know point A and B clearly.
Now my issue is the following:
i would like to shorten both ends of the line but maintain its angle.
I tried just subtracting manually a value from y position but its not always accurate. see the thing is the points i will be given ALWAYS make the line point to the center of the markers on the map. See how in the photo the line points directly to the center of the "B" marker. so i just need to maintain its angle and i'll be fine. So how can i shorten the line on both ends but still maintain the correct angle ?
the coordinate system i am using looks like this:
arithmetic graphing-functions slope
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closed as unclear what you're asking by Eevee Trainer, Lord Shark the Unknown, José Carlos Santos, mrtaurho, metamorphy Jan 1 at 14:24
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have a line that looks like this:
to draw this line i have the following:
point A: (x,y)
point B: (x,y)
so for example, i will be given this path:
(90,20),(400,300)
and then i can draw a line because i know point A and B clearly.
Now my issue is the following:
i would like to shorten both ends of the line but maintain its angle.
I tried just subtracting manually a value from y position but its not always accurate. see the thing is the points i will be given ALWAYS make the line point to the center of the markers on the map. See how in the photo the line points directly to the center of the "B" marker. so i just need to maintain its angle and i'll be fine. So how can i shorten the line on both ends but still maintain the correct angle ?
the coordinate system i am using looks like this:
arithmetic graphing-functions slope
$endgroup$
closed as unclear what you're asking by Eevee Trainer, Lord Shark the Unknown, José Carlos Santos, mrtaurho, metamorphy Jan 1 at 14:24
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hi & welcome to MSE. Here is a hint. Equation of a Line from 2 Points shows how to create a simple equation describing the line. You can then adjust the value of $x$ or $y$ to determine the other value, i.e., $y$ or $x$.
$endgroup$
– John Omielan
Jan 1 at 4:03
$begingroup$
Do you want to shorten both ends equally, or one more than the other? How are you given the amount to shorten?
$endgroup$
– Ross Millikan
Jan 1 at 4:19
$begingroup$
yes. i just want to shorten it slightly. i dont have an exact number . i'll play with it. can you help ?
$endgroup$
– j2emanue
Jan 1 at 4:27
add a comment |
$begingroup$
I have a line that looks like this:
to draw this line i have the following:
point A: (x,y)
point B: (x,y)
so for example, i will be given this path:
(90,20),(400,300)
and then i can draw a line because i know point A and B clearly.
Now my issue is the following:
i would like to shorten both ends of the line but maintain its angle.
I tried just subtracting manually a value from y position but its not always accurate. see the thing is the points i will be given ALWAYS make the line point to the center of the markers on the map. See how in the photo the line points directly to the center of the "B" marker. so i just need to maintain its angle and i'll be fine. So how can i shorten the line on both ends but still maintain the correct angle ?
the coordinate system i am using looks like this:
arithmetic graphing-functions slope
$endgroup$
I have a line that looks like this:
to draw this line i have the following:
point A: (x,y)
point B: (x,y)
so for example, i will be given this path:
(90,20),(400,300)
and then i can draw a line because i know point A and B clearly.
Now my issue is the following:
i would like to shorten both ends of the line but maintain its angle.
I tried just subtracting manually a value from y position but its not always accurate. see the thing is the points i will be given ALWAYS make the line point to the center of the markers on the map. See how in the photo the line points directly to the center of the "B" marker. so i just need to maintain its angle and i'll be fine. So how can i shorten the line on both ends but still maintain the correct angle ?
the coordinate system i am using looks like this:
arithmetic graphing-functions slope
arithmetic graphing-functions slope
edited Jan 1 at 4:02
Gerry Myerson
147k8151304
147k8151304
asked Jan 1 at 3:59
j2emanuej2emanue
1064
1064
closed as unclear what you're asking by Eevee Trainer, Lord Shark the Unknown, José Carlos Santos, mrtaurho, metamorphy Jan 1 at 14:24
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Eevee Trainer, Lord Shark the Unknown, José Carlos Santos, mrtaurho, metamorphy Jan 1 at 14:24
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hi & welcome to MSE. Here is a hint. Equation of a Line from 2 Points shows how to create a simple equation describing the line. You can then adjust the value of $x$ or $y$ to determine the other value, i.e., $y$ or $x$.
$endgroup$
– John Omielan
Jan 1 at 4:03
$begingroup$
Do you want to shorten both ends equally, or one more than the other? How are you given the amount to shorten?
$endgroup$
– Ross Millikan
Jan 1 at 4:19
$begingroup$
yes. i just want to shorten it slightly. i dont have an exact number . i'll play with it. can you help ?
$endgroup$
– j2emanue
Jan 1 at 4:27
add a comment |
1
$begingroup$
Hi & welcome to MSE. Here is a hint. Equation of a Line from 2 Points shows how to create a simple equation describing the line. You can then adjust the value of $x$ or $y$ to determine the other value, i.e., $y$ or $x$.
$endgroup$
– John Omielan
Jan 1 at 4:03
$begingroup$
Do you want to shorten both ends equally, or one more than the other? How are you given the amount to shorten?
$endgroup$
– Ross Millikan
Jan 1 at 4:19
$begingroup$
yes. i just want to shorten it slightly. i dont have an exact number . i'll play with it. can you help ?
$endgroup$
– j2emanue
Jan 1 at 4:27
1
1
$begingroup$
Hi & welcome to MSE. Here is a hint. Equation of a Line from 2 Points shows how to create a simple equation describing the line. You can then adjust the value of $x$ or $y$ to determine the other value, i.e., $y$ or $x$.
$endgroup$
– John Omielan
Jan 1 at 4:03
$begingroup$
Hi & welcome to MSE. Here is a hint. Equation of a Line from 2 Points shows how to create a simple equation describing the line. You can then adjust the value of $x$ or $y$ to determine the other value, i.e., $y$ or $x$.
$endgroup$
– John Omielan
Jan 1 at 4:03
$begingroup$
Do you want to shorten both ends equally, or one more than the other? How are you given the amount to shorten?
$endgroup$
– Ross Millikan
Jan 1 at 4:19
$begingroup$
Do you want to shorten both ends equally, or one more than the other? How are you given the amount to shorten?
$endgroup$
– Ross Millikan
Jan 1 at 4:19
$begingroup$
yes. i just want to shorten it slightly. i dont have an exact number . i'll play with it. can you help ?
$endgroup$
– j2emanue
Jan 1 at 4:27
$begingroup$
yes. i just want to shorten it slightly. i dont have an exact number . i'll play with it. can you help ?
$endgroup$
– j2emanue
Jan 1 at 4:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the coordinates of $A$ are $(Ax,Ay)$ and of $B$ are $(Bx,By)$ the current length of the line is $$L=sqrt{(Ax-Bx)^2+(Ay-By)^2}$$
You can parameterize the line as
$$(Ax+t(Bx-Ax),Ay+t(By-Ay))$$
When $t=0$ you are at $A$ and when $t=1$ you are at $B$. If you want to shorten the line, just go from, say, $t=0.1$ to $t=0.8$. That will shorten the line by $10%$ at the $A$ end and $20%$ at the $B$ end.
$endgroup$
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the coordinates of $A$ are $(Ax,Ay)$ and of $B$ are $(Bx,By)$ the current length of the line is $$L=sqrt{(Ax-Bx)^2+(Ay-By)^2}$$
You can parameterize the line as
$$(Ax+t(Bx-Ax),Ay+t(By-Ay))$$
When $t=0$ you are at $A$ and when $t=1$ you are at $B$. If you want to shorten the line, just go from, say, $t=0.1$ to $t=0.8$. That will shorten the line by $10%$ at the $A$ end and $20%$ at the $B$ end.
$endgroup$
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
add a comment |
$begingroup$
If the coordinates of $A$ are $(Ax,Ay)$ and of $B$ are $(Bx,By)$ the current length of the line is $$L=sqrt{(Ax-Bx)^2+(Ay-By)^2}$$
You can parameterize the line as
$$(Ax+t(Bx-Ax),Ay+t(By-Ay))$$
When $t=0$ you are at $A$ and when $t=1$ you are at $B$. If you want to shorten the line, just go from, say, $t=0.1$ to $t=0.8$. That will shorten the line by $10%$ at the $A$ end and $20%$ at the $B$ end.
$endgroup$
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
add a comment |
$begingroup$
If the coordinates of $A$ are $(Ax,Ay)$ and of $B$ are $(Bx,By)$ the current length of the line is $$L=sqrt{(Ax-Bx)^2+(Ay-By)^2}$$
You can parameterize the line as
$$(Ax+t(Bx-Ax),Ay+t(By-Ay))$$
When $t=0$ you are at $A$ and when $t=1$ you are at $B$. If you want to shorten the line, just go from, say, $t=0.1$ to $t=0.8$. That will shorten the line by $10%$ at the $A$ end and $20%$ at the $B$ end.
$endgroup$
If the coordinates of $A$ are $(Ax,Ay)$ and of $B$ are $(Bx,By)$ the current length of the line is $$L=sqrt{(Ax-Bx)^2+(Ay-By)^2}$$
You can parameterize the line as
$$(Ax+t(Bx-Ax),Ay+t(By-Ay))$$
When $t=0$ you are at $A$ and when $t=1$ you are at $B$. If you want to shorten the line, just go from, say, $t=0.1$ to $t=0.8$. That will shorten the line by $10%$ at the $A$ end and $20%$ at the $B$ end.
answered Jan 1 at 4:35
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
add a comment |
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
i need the new coordinates to plot. this will give me the length. i cannot use that.
$endgroup$
– j2emanue
Jan 1 at 4:38
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
$begingroup$
You can compute them from whatever $t$ values you pick for the ends of the line. Just plug those $t$s into the second expression.
$endgroup$
– Ross Millikan
Jan 1 at 4:48
add a comment |
1
$begingroup$
Hi & welcome to MSE. Here is a hint. Equation of a Line from 2 Points shows how to create a simple equation describing the line. You can then adjust the value of $x$ or $y$ to determine the other value, i.e., $y$ or $x$.
$endgroup$
– John Omielan
Jan 1 at 4:03
$begingroup$
Do you want to shorten both ends equally, or one more than the other? How are you given the amount to shorten?
$endgroup$
– Ross Millikan
Jan 1 at 4:19
$begingroup$
yes. i just want to shorten it slightly. i dont have an exact number . i'll play with it. can you help ?
$endgroup$
– j2emanue
Jan 1 at 4:27